chapter find standard form. then factor and use the principle of zero products

CHAPTER

11 Quadratic Equations and Functions

Slide 2 Copyright 2011, 2007, 2003, 1999 Pearson Education, Inc.

11.1 The Basics of Solving Quadratic Equations 11.2 The Quadratic Formula 11.3 Applications Involving Quadratic Equations 11.4 More on Quadratic Equations 11.5 Graphing f(x) = a(x – h)2 + k 11.6 Graphing f(x) = ax2 + bx + c 11.7 Mathematical Modeling with Quadratic Functions 11.8 Polynomial Inequalities and Rational Inequalities

OBJECTIVES

11.1 The Basics of Solving Quadratic Equations


a Solve quadratic equations using the principle of square roots and find the x-intercepts of the graph of a related function.

b Solve quadratic equations by completing the square. c Solve applied problems using quadratic equations.


Quadratic Equation


An equation of the type ax2 + bx + c = 0, where a, b, and c are real-number constants and a > 0, is called the standard form of a quadratic equation.

EXAMPLE


a Solve quadratic equations using the principle of square roots.

2 Solve.


EXAMPLE Solution


a Solve quadratic equations using the principle of square roots.

2


Factor and use the principle of zero products:

EXAMPLE


a Solve quadratic equations and find the x-intercepts of the graph of a related function.

3 Solve and find the x-intercepts.


EXAMPLE Solution



3


a) First find standard form. Then factor and use the principle of zero products.

EXAMPLE Solution



3


EXAMPLE Solution

b) The x-intercepts of f(x) = 3x2 + x – 2 are

and The solutions of the equation 3x2 = 2 – x are the first coordinates of the x-intercepts of the graph of

f(x) = 3x2 + x – 2.



3


The solutions of the equation x2 = d are and – . When d > 0, the solutions are two real numbers. When d = 0, the only solution is 0. When d < 0, the solutions are two imaginary numbers.


The Principle of Square Roots


EXAMPLE


a Solve quadratic equations.

4 Solve.


Give the exact solutions and approximate the solutions to three decimal places.

EXAMPLE Solution

The symbol is often used to represent both of the solutions.


a Solve quadratic equations..

4


EXAMPLE Solution



4


EXAMPLE



6 Solve.


EXAMPLE Solution



6


EXAMPLE



8 Solve.


EXAMPLE Solution



8



Completing the Square


When solving an equation, to complete the square of an expression like x2 + bx, take half the x-coefficient, which is b/2, and square it. Then we add that number, (b/2)2, on both sides of the equation.

EXAMPLE


b Solve quadratic equations by completing the square.

10 Solve by completing the square.


EXAMPLE Solution



10


EXAMPLE



12 Solve by completing the square.


EXAMPLE Solution



12



Solve by Completing the Square


To solve an equation ax2 + bx + c = 0 by completing the square: 1. If a ≠ 1, multiply by 1/a so that the x2-coefficient is 1. 2. If the x2-coefficient is 1, add or subtract so that the

equation is in the form if step (1) has been applied.


Solve by Completing the Square


3. Take half of the x-coefficient and square it. Add the result on both sides of the equation.

4. Express the side with the variables as the square of a binomial.

5. Use the principle of square roots and complete the solution.

EXAMPLE


c Solve applied problems using quadratic equations.

13 Hang Time.


One of the most exciting plays in basketball is the dunk shot. The amount of time T that passes from the moment a player leaves the ground, goes up, makes the shot, and arrives back on the ground is called hang time. A function relating an athlete’s vertical leap V, in inches, to hang time T, in seconds, is given by V(T) = 48T2.

EXAMPLE



13 Hang Time.


a) Hall-of-Famer Michael Jordan had a hang time of about 0.889 sec. What was his vertical leap?

b) Although his height is only 5 ft 7 in., Spud Webb, formerly of the Sacramento Kings, had a vertical leap of about 44 in. What was his hang time?

EXAMPLE Solution



13


a) To find Jordan’s vertical leap, we substitute 0.889 for T in the function and compute V:

Jordan’s vertical leap was about 37.9 in.

b) To find Webb’s hang time, substitute 44 for V and solve for T:

EXAMPLE Solution Webb’s hang time was 0.957 sec.



13


CHAPTER




OBJECTIVES

11.2 The Quadratic Formula


a Solve quadratic equations using the quadratic formula, and approximate solutions using a calculator.


The Quadratic Formula




To solve a quadratic equation: 1. Check for the form x2 = d or (x + c)2 = d. If it is in this

form, use the principle of square roots. 2. If it is not in the form of step (1), write it in standard

form ax2 + bx + c = 0 with a and b nonzero. 3. Then try factoring.



4. If it is not possible to factor or if factoring seems difficult, use the quadratic formula. The solutions of a quadratic equation cannot always be found by factoring. They can always be found using the quadratic formula.

EXAMPLE



2 Solve.


Give the exact solutions and approximate the solutions to three decimal places.

EXAMPLE Solution



2


First find standard form and determine a, b, and c.

EXAMPLE Solution



2


Then use the quadratic formula

EXAMPLE Solution



2


Use a calculator to approximate the solutions:

The check for is left to the student.

EXAMPLE



4 Solve.


Give the exact solutions and approximate solutions to three decimal places.

EXAMPLE Solution



4


First find an equivalent quadratic equation in standard form:

Then

EXAMPLE Solution



4


EXAMPLE Solution



4


Use a calculator to approximate the solutions:

The solutions are

CHAPTER




OBJECTIVES

11.3 Applications Involving Quadratic Equations


a Solve applied problems involving quadratic equations. b Solve a formula for a given letter.

EXAMPLE


a Solve applied problems involving quadratic equations.

1 Quilt Dimensions.


Michelle is making a quilt for a wall hanging at the entrance of a state museum. The finished quilt will measure 8 ft by 6 ft. The quilt has a border of uniform width around it. The area of the interior rectangular section is one-half the area of the entire quilt. How wide is the border?

EXAMPLE Solution



1


1. Familiarize. We don’t know how wide the border is, so we have called its width x.

2. Translate. Remember, the area of a rectangle is lw (length times width).

Then:

EXAMPLE Solution



1


Since the area of the interior section is one-half the area of the entire quilt,

EXAMPLE Solution



1


3. Solve.

EXAMPLE Solution



1


4. Check. We check in the original problem. We see that 6 is not a solution because an 8-ft by 6-ft quilt cannot have a 6-ft border. When x = 6, then 8 – 2x = –4 and

d – 2x = –6 and the dimensions of the interior section of the quilt cannot be negative.

EXAMPLE Solution



1


If the border is 1 ft wide, then the interior will have length 8 – 2 1, or 6 ft. The width will be 6 – 2 1, or 4 ft. The area of the interior is thus 6 4, or 24 ft2. The area of the entire quilt is 8 6, or 48 ft2. The area of the interior is one-half of 48 ft2 so the number 1 checks.

5. State. The border of the quilt is 1 ft wide.

EXAMPLE



2 Town Planning.


Three towns A, B, and C are situated. The roads at A form a right angle. The distance from A to B is 2 mi less than the distance from A to C. The distance from B to C is 10 mi. Find the distance from A to B and the distance from A to C.

EXAMPLE



2 Town Planning.


EXAMPLE Solution



2


1. Familiarize. First make a drawing and label it. Let the distance from A to C. Then the distance from A to B is d – 2.

EXAMPLE Solution



2


2. Translate.

3. Solve.

EXAMPLE Solution



2


4. Check. –6 cannot be a solution because distances are not negative. If d = 8 then d – 2 = 6 and

EXAMPLE Solution



2


Since 102 = 100, the distance 8 mi checks. 5. State. The distance from A to C is 8 mi, and the

distance from A to B is 6 mi.

EXAMPLE



4 Motorcycle Travel.


Karin’s motorcycle traveled 300 mi at a certain speed. Had she gone 10 mph faster, she could have made the trip in 1 hr less time. Find her speed.

EXAMPLE Solution



4


1. Familiarize.

Recalling the motion formula d = rt and solving for r, get r = d/t. From the rows of the table,

EXAMPLE Solution



4


2. Translate. Substitute for r from the first equation into the second and get a translation:

3. Solve.

EXAMPLE Solution



4


EXAMPLE Solution



4


4. Check. Since negative time has no meaning in this problem, we try 6 hr. Remembering that r = d/t, we get

r = 300/6 = 50 mph. To check, we take the speed 10 mph faster, which is 60 mph, and see how long the trip would have taken at that speed:

EXAMPLE Solution



4


This is 1 hr less than the trip actually took, so we have an answer.

5. State. Karin’s speed was 50 mph.

EXAMPLE


b Solve a formula for a given letter.

5 Period of a Pendulum.


The time T required for a pendulum of length L to swing back and forth (complete one period) is given by the formula where g is the gravitational constant. Solve for L.

EXAMPLE Solution



5


EXAMPLE

An object that is tossed downward with an initial speed (velocity) of will travel a distance of meters, where and t is measured in seconds. Solve for t.



7 Falling Distance.


EXAMPLE Solution



7


Since t is squared in one term and raised to the first power in the other term, the equation is quadratic in t. The variable is and s are treated as constants.

EXAMPLE Solution



7


EXAMPLE Solution



7


Use only the positive root:

EXAMPLE



8 Solve for a.


EXAMPLE Solution



8


Clear the fractions. Multiplying by we have

Now square both sides.

EXAMPLE Solution



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CHAPTER




OBJECTIVES

11.4 More on Quadratic Equations


a Determine the nature of the solutions of a quadratic equation.

b Write a quadratic equation having two given numbers as solutions.

c Solve equations that are quadratic in form.




The solutions x1 and x2 of a quadratic equation are given by




The expression is called the discriminant.

EXAMPLE



2 Determine the nature of the solutions.


EXAMPLE Solution



2


Since the discriminant is negative, there are two nonreal complex-number solutions.

EXAMPLE Solution



2


EXAMPLE



3 Determine the nature of the solutions.


EXAMPLE Solution



3


Since the discriminant is positive, there are two solutions, and they are real numbers. The equation can be solved by factoring since the discriminant is a perfect square.

EXAMPLE Solution



3


EXAMPLE



5 Find the quadratic equation.


Find a quadratic equation whose solutions are 3 and

EXAMPLE Solution



5


EXAMPLE



7 Write a quadratic equation.


Write a quadratic equation whose solutions are and

EXAMPLE Solution



7


EXAMPLE



8 Write a quadratic equation.


Write a quadratic equation whose solutions are –12i and 12i .

EXAMPLE Solution



8





Certain equations that are not really quadratic can still be solved as quadratic. Equations that can be solved like this are said to be quadratic in form, or reducible to quadratic.

EXAMPLE



10 Solve.


EXAMPLE Solution



10


EXAMPLE Solution



10


Next, substitute for and solve these equations:

Squaring the first equation, get x = 16. Squaring the second equation, get x = 1. Check both solutions.

EXAMPLE Solution



10


Since 16 checks but 1 does not, the solution is 16.

EXAMPLE



12 Find the x-intercepts of the graph.


EXAMPLE Solution



12


The x-intercepts occur where f(x) = 0, so we must have

Let u = x2 – 1. Then solve the equation found by substituting u for x2 – 1.

EXAMPLE Solution



12


Next, substitute x2 – 1 for u and solve these equations:

EXAMPLE Solution



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CHAPTER




OBJECTIVES

11.5 Graphing f(x) = a(x – h)2 + k


a Graph quadratic functions of the type f(x) = ax2 and then label the vertex and the line of symmetry.

b Graph quadratic functions of the type f(x) = a(x – h)2 and then label the vertex and the line of symmetry.

c Graph quadratic functions of the type f(x) = a(x – h)2+k, finding the vertex, the line of symmetry, and the maximum or minimum function value, or y-value.

EXAMPLE



1 Graph.


EXAMPLE Solution



1


Choose some values for x and compute f(x) for each. Then plot the ordered pairs and connect them with a smooth curve.

EXAMPLE Solution



1


EXAMPLE Solution



1


All quadratic function graphs have curves called parabolas. They are cup-shaped curves that are symmetric with respect to a vertical line known as the parabola’s line of symmetry, or axis of symmetry. The y-axis (or the line x = 0) is the line of symmetry. If the paper were to be folded on this line, the two halves of the curve would coincide.


Graphs of f(x) = ax2


EXAMPLE



2 Graph.


EXAMPLE Solution



2


Choose some values for x and compute g(x). Then plot the points and draw the curve.

EXAMPLE Solution



2



Graphs of f(x) = a(x – h)2


The graph of f(x) = a(x – h)2 has the same shape as the graph of y = ax2. If h is positive, the graph of y = ax2 is shifted h units to the right. If h is negative, the graph of y = ax2 is shifted units to the left. The vertex is (h, 0), and the line of symmetry is x = h.

EXAMPLE



4 Graph.


EXAMPLE Solution First rewrite the equation as In this case, a = –2 and h = –3, so the graph looks like that of g(x) = 2x2 translated 3 units to the left and, since –2 < 0, the graph opens down. The vertex is (–3, 0), and the line of symmetry is x = –3. Plot points as needed.



4


EXAMPLE Solution



4



c Graph quadratic functions of the type f(x) = a(x – h)2+k, finding the vertex, the line of symmetry, and the maximum or minimum function value, or y-value.


Note that if a parabola opens up (a > 0), the function value, or y-value, at the vertex is a least, or minimum, value. That is, it is less than the y-value at any other point on the graph. If the parabola opens down (a < 0 ), the function value at the vertex is a greatest, or maximum, value.


Graphs of f(x) = a(x – h)2 + k


The graph of f(x) = a(x – h)2 + k has the same shape as the graph of y = a(x – h)2. If k is positive, the graph of y = a(x – h)2 is shifted k units up. If k is negative, the graph of y = a(x – h)2 is shifted down. The vertex is (h, k) and the line of symmetry is x = h. For a > 0 , k is the minimum function value. For a < 0, k is the maximum function value.

EXAMPLE


c Graph quadratic functions, finding the vertex, line of symmetry, and maximum or minimum function value.

7 Graph.


Find the vertex, the line of symmetry, and the maximum or minimum value.

EXAMPLE Solution



7


First express the equation in the equivalent form

The graph looks like that of g(x) = –2x2 translated 3 units to the left and 5 units up. The vertex is (–3, 5), and the line of symmetry is x = –3. Since –2 < 0, we know that the graph opens down so 5, the second coordinate of the vertex, is the maximum y-value.

EXAMPLE Solution



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CHAPTER




OBJECTIVES

11.6 Graphing f(x) = ax2 + bx + c


a For a quadratic function, find the vertex, the line of symmetry, and the maximum or minimum value, and then graph the function.

b Find the intercepts of a quadratic function.

EXAMPLE


a Graph a quadratic function, then find the vertex, line of symmetry, and maximum or minimum value.

1 Graph.


For f(x) = x2 – 6x + 4, find the vertex, the line of symmetry, and the maximum or the minimum value. Then graph.

EXAMPLE Solution



1


First find the vertex and the line of symmetry. To do so, find the equivalent form by completing the square.

Complete the square inside the parentheses, but in a different manner than before. Take half the x-coefficient, and square it: (–3)2 = 9. Then add 0, or 9 – 9 inside the parentheses.

EXAMPLE Solution



1


EXAMPLE Solution



1


The vertex is (3, –5), and the line of symmetry is x = 3. The coefficient of x2 is 1, which is positive, so the graph opens up. This tells us that –5 is the minimum value. Plot the vertex and draw the line of symmetry. Choose some x-values on both sides of the vertex and graph the parabola. Compute the pair (5, –1).

EXAMPLE Solution



1


Note that it is 2 units to the right of the line of symmetry. There will also be a pair with the same y-coordinate on the graph 2 units to the left of the line of symmetry. Thus we get a second point, (1, –1), without making another calculation

EXAMPLE Solution



1


EXAMPLE



3 Graph.


For f(x) = –2x2 + 10x – 7, find the vertex, the line of symmetry, and the maximum or the minimum value. Then graph.

EXAMPLE Solution



3


The coefficient of x2 is not 1. Factor out –2 from only the first two terms of the expression. This makes the coefficient of x2 inside the parentheses 1.

EXAMPLE Solution



3


Next, complete the square as before.

EXAMPLE Solution



3


The vertex is and the line of symmetry is The coefficient of x2 is –2 so the graph is narrow and opens down. This tells us that is the maximum value of the function.

EXAMPLE Solution



3


Choose a few x-values on one side of the line of symmetry, compute y-values, and use the resulting coordinates to find more points on the other side of the line of symmetry. Plot points and graph the parabola.

EXAMPLE Solution



3


The vertex of a parabola given by f(x) = ax2 + bx + c is The x-coordinate of the vertex is –b/(2a). The line of symmetry is x = –b/(2a). The second coordinate of the vertex is easiest to find by computing


Vertex; Line of Symmetry


EXAMPLE



4 Find the intercepts.


EXAMPLE Solution

The y-intercept is (0, f(0)). Since the y-intercept is (0, –2). To find the x-intercepts, solve:



4


EXAMPLE Solution



4


Using the quadratic formula, we have Thus the x–intercepts are approximately, (–0.732, 0) and (2.732, 0).

EXAMPLE Solution



4


CHAPTER




OBJECTIVES

11.7 Mathematical Modeling with Quadratic Functions


a Solve maximum–minimum problems involving quadratic functions.

b Fit a quadratic function to a set of data to form a mathematical model, and solve related applied problems.

EXAMPLE



1 Bordered Garden.


Millie is planting a garden to produce vegetables and fruit for the local food bank. She has enough raspberry plants to edge a 64-yd perimeter and wants to maximize the area within to plant the most vegetables possible. What are the dimensions of the largest rectangular garden that Millie can enclose with the raspberry plants?

EXAMPLE Solution



1


1. Familiarize. Let l = the length of the garden and w = the width. Recall the following formulas:

Perimeter: 2l + 2w; Area: l w.

2. Translate.

EXAMPLE Solution



1


To express A in terms of w, solve for l in the first equation:

EXAMPLE Solution



1


Substituting 32 – w for l, we get a quadratic function A(w), or just A.

EXAMPLE Solution



1


3. Carry out.

EXAMPLE Solution



1


The vertex is (16, 256). Thus the maximum value is 256. It occurs when w = 16 and l = 32 – w = 32 – 16 = 16.

4. Check. Note that 256 is larger than any of the values found in the Familiarize step.

5. State. The largest rectangular garden that can be enclosed is 16 yd by 16 yd; that is, it is a square with sides of 16 ft.


b Fit a quadratic function to a set of data.


EXAMPLE


b Fit a quadratic function to a set of data to form a mathematical model.

Choosing models.


For the scatterplots , determine which, if any, of the following functions might be used as a model for the data. Linear, Quadratic, Quadratic, Polynomial, neither quadratic nor linear

EXAMPLE



Choosing models.


EXAMPLE Solution




The data rise and then fall in a curved manner fitting a quadratic function

EXAMPLE Solution




EXAMPLE


b Fit a quadratic function to a set of data to solve related applied problems.

7 Canoe Depth.


The drawing shows the cross section of a canoe. Canoes are deepest at the middle of the center line, with the depth decreasing to zero at the edges. Lou and Jen own a company that specializes in producing custom canoes. A customer provided suggested guidelines for measures of the depths D, in inches, along the center line of the canoe at distances x, in inches, from the edge. The measures are listed in the table.

EXAMPLE



7 Canoe Depth.


EXAMPLE



7 Canoe Depth.


a) Make a scatterplot of the data. b) Decide whether the data seem to fit a quadratic

function. c) Use the data points (0, 0), (18, 14), and (36, 0) to find a

quadratic function that fits the data. d) Use the function to estimate the depth of the canoe

10 in. from the edge along the center line.

EXAMPLE Solution


b Fit a quadratic function to a set of data to solve related applied problems..

7


a) The red squares comprise the scatterplot.

EXAMPLE Solution



7


b) The data seem to rise and fall in a manner similar to a quadratic function. The dashed black line in the graph represents a sample quadratic function of fit. Note that it may not necessarily go through each point.

c) We are looking for a quadratic function

EXAMPLE Solution



7


Determine the constants a, b, and c. Use the three data points (0, 0), (18, 14), and (36, 0) and substitute as follows:

EXAMPLE Solution



7


After simplifying, solve the system.

Since c = 0, the system reduces to a system of two equations in two variables:

EXAMPLE Solution



7


Multiply equation (1) by –2, add, and solve for a.

EXAMPLE Solution



7


Next, substitute for a in equation (2) and solve for b.

EXAMPLE Solution



7


This gives us the quadratic function:

EXAMPLE Solution



7


d) To find the depth 10 in. from the edge of the canoe, substitute:

At a distance of 10 in. from the edge of the canoe, the depth of the canoe is about 11.26 in.

EXAMPLE Solution



7


CHAPTER




OBJECTIVES

11.8 Polynomial Inequalities and Rational Inequalities


a Solve quadratic inequalities and other polynomial inequalities.

b Solve rational inequalities.




Inequalities like the following are called quadratic inequalities:

EXAMPLE



1 Solve.


EXAMPLE Solution



1


Consider the function f(x) = x2 + 3x –10 and its graph. The graph opens up since the leading coefficient (a = 1) is positive. Find the x-intercepts by setting the polynomial equal to 0 and solving:

EXAMPLE Solution



1


EXAMPLE Solution



1


Values of will be positive to the left and right of the intercepts. Thus the solution set of the inequality is




To solve a polynomial inequality: 1. Get 0 on one side, set the expression on the other side

equal to 0, and solve to find the x-intercepts. 2. Use the numbers found in step (1) to divide the number

line into intervals.




3. Substitute a number from each interval into the related function. If the function value is positive, then the expression will be positive for all numbers in the interval. If the function value is negative, then the expression will be negative for all numbers in the interval.

4. Select the intervals for which the inequality is satisfied and write set-builder notation or interval notation for the solution set.

EXAMPLE



4 Solve.


EXAMPLE Solution



4


The solutions of f(x) = 0, or 5x(x + 3)(x – 2) = 0 are 0, –3, and 2. They divide the real-number line into four intervals, as shown below.

EXAMPLE Solution



4


Try test numbers in each interval:

EXAMPLE Solution



4


EXAMPLE Solution

The expression is positive for values of x in intervals B and D. Since the inequality symbol is we need to include the x-intercepts. The solution set of the inequality is



4


EXAMPLE Solution



4





Rational inequalities are inequalities that involve rational expressions.

EXAMPLE



5 Solve.


EXAMPLE Solution



5


Write a related equation by changing the symbol to =.

Solve this related equation.

EXAMPLE Solution



5


With rational inequalities, we also need to determine those numbers for which the rational expression is not defined—that is, those numbers that make the denominator 0. Set the denominator equal to 0 and solve: x + 4 = 0, or x = –4. Next, use the numbers –11 and –4 to divide the number line into intervals.

EXAMPLE Solution



5


Since the inequality is false for x = –15, the number –15 is not a solution of the inequality. Interval A is not part of the solution set.

EXAMPLE Solution



5


EXAMPLE Solution

The solution set includes the interval B. The number –11 is also included since the inequality symbol is and –11 is a solution of the related equation. The number –4 is not included; it is not an allowable replacement because it results in division by 0.



5


EXAMPLE Solution



5


Thus the solution set of the original inequality is

chapter find standard form. then factor and use the principle of zero products

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